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ASHA 777 [7]
3 years ago
13

The equilibrium of 2H 2 O(g) 2H 2 (g) + O 2 (g) at 2,000 K has a Keq value of 5.31 x 10-10. What is the Keq expression for this

system? (Enter subscripts after the letters: for example, H2O = Hs2O. Also, dont forget to use proper chemical shorthand for chemical symbols. For instance, chlorine = Cl but not cl or cL.)
Chemistry
2 answers:
melomori [17]3 years ago
4 0

Answer : The equilibrium constant expression will be,

k_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}

Explanation :

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given balanced chemical reaction is,

2H_2O(g)\rightleftharpoons 2H_2(g)+O_2(g)

So, the equilibrium constant expression will be,

k_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}

nevsk [136]3 years ago
3 0

Answer:

5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}

Explanation:

For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.

For a hypothetical reaction:

xA + yB ⇄ zC

The equilibrium constant is :

Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }

The given reaction involves the decomposition of H2O into H2 and O2

2H_{2}O\rightleftharpoons 2H_{2} + O_{2}

The equilibrium constant is expressed as :

Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}

Since Keq = 5.31*10^-10

5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}

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The three reactions involved in the Ostwald process for the conversion of NH3 to HNO3 are:

4NH3(g) + 5O2(g) ==> 4NO(g) + 6H2O(l)   ------(1)

2NO(g) + O2(g) ==> 2NO2(g)  ------------------------(2)

3NO2(g) + H2O(l) ==> 2HNO3(aq) + NO(g) ---(3)

Mass of HNO3 produced = 1.81 tons

In grams, the mass of HNO3 = 1.81*2000*453.6 = 1642032 g

Molar mass of HNO3 = 63 g/mol

Thus, # moles of HNO3 produced = 1642032/63 = 26064 moles

Based on the stoichiometry of reaction (3):

Theoretical moles of NO2 = 1.5(Moles HNO2 ) = 1.5(26064) = 39096

Assuming 80% yield:

Actual moles of NO2 = 0.8*39096 = 31277 moles

Based on the stoichiometry in reaction (2):

Theoretical moles of NO = 31277 moles

Assuming 80% yield:

Actual moles of NO = 0.8*31277 = 25022 moles

Bases on stoichiometry in reaction(3):

Moles of NH3 = 25022 moles

Assuming 80% yield:

Actual moles of NH3 required = 0.8*25022 = 20018 moles

Mass of NH3 required = 20018 moles * 17 g/mole = 340306 g = 340.31 kg

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The balanced equation  

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Number of moles of A = Number of moles of HA  = Number of moles of NaOH

= 35.8/1000 * 0.020 = 0.000716 mol

Initial concentration of A = 0.000716/0.0608 = 0.01178 M

pKb = 14 – pKa = 14 -3.9 = 10.1

Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11

Kb = [HA][OH-]/[A-]

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pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02

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